The number of real values of $$k$$ for which the lines $$\frac{{x - k}}{4} = \frac{{y - 1}}{2} = \frac{{z + 1}}{1}$$     and $$\frac{{x - \left( {k + 1} \right)}}{1} = \frac{y}{{ - 1}} = \frac{{z - 1}}{2}$$      are intersecting, is :

Solution :
Any point on the first line is $$\left( {4r + k,\,2r + 1,\,r - 1} \right),$$     and any point on the second line is $$\left( {r' + k + 1,\, - r',\,2r' + 1} \right).$$      The lines are intersecting if $$4r + k = r' + k + 1,\,2r + 1 = - r',\,r - 1 = 2r' + 1$$          for some $$r$$ and $$r' \Rightarrow 4r - r' = 1,\,2r + r' = - 1,\,r - 2r' = 2.$$
Now, $$4r - r' = 1,\,2r + r' = - 1\,\, \Rightarrow r = 0,\,r' = - 1$$         which satisfy $$r - 2r' = 2.$$
This is true for all $$k.$$